Difference of Two Powers/Proof 1

Theorem

Let $\mathbb F$ denote one of the standard number systems, that is $\Z$, $\Q$, $\R$ and $\C$.

Let $n \in \N$ such that $n \ge 2$.

Then for all $a, b \in \mathbb F$:

\(\ds a^n - b^n\)

\(=\)

\(\ds \paren {a - b} \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j\)

\(\ds \)

\(=\)

\(\ds \paren {a - b} \paren {a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \dotsb + a b^{n - 2} + b^{n - 1} }\)

Proof

Let $\ds S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$.

This can also be written:

$\ds S_n = \sum_{j \mathop = 0}^{n - 1} b^j a^{n - j - 1}$

Consider:

$\ds a S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} b^j$

Taking the first term (where $j = 0$) out of the summation, we get:

$\ds a S_n = \sum_{j \mathop = 0}^{n - 1} a^{n - j} b^j = a^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j$

Similarly, consider:

$\ds b S_n = \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j}$

Taking the first term (where $j = 0$) out of the summation:

$\ds b S_n = \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j} = b^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j$

This is equal to:

$\ds b^n + \sum_{j \mathop = 1}^{n - 1} a^j b^{n - j}$

by Permutation of Indices of Summation.

So:

\(\ds \paren {a - b} S_n\)

\(=\)

\(\ds a \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j - b \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j - 1}\)

\(\ds \)

\(=\)

\(\ds \sum_{j \mathop = 0}^{n - 1} a^{n - j} b^j - \sum_{j \mathop = 0}^{n - 1} a^j b^{n - j}\)

\(\ds \)

\(=\)

\(\ds a^n + \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j - \sum_{j \mathop = 1}^{n - 1} a^{n - j} b^j - b^n\)

\(\ds \)

\(=\)

\(\ds a^n - b^n\)

$\blacksquare$